document.write( "Question 446629: Their is a rectangle and its legnth is d-12 and the width is d. If the area is 28 square units, what is the value of d? \n" ); document.write( "
Algebra.Com's Answer #307581 by chriswen(106)\"\" \"About 
You can put this solution on YOUR website!
A=l*w
\n" ); document.write( "28=(d)(d-12)
\n" ); document.write( "28=d^2-12d
\n" ); document.write( "d^2-12d-28=0
\n" ); document.write( "\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( "
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation \"ad%5E2%2Bbd%2Bc=0\" (in our case \"1d%5E2%2B-12d%2B-28+=+0\") has the following solutons:
\n" ); document.write( "
\n" ); document.write( " \"d%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca\"
\n" ); document.write( "
\n" ); document.write( " For these solutions to exist, the discriminant \"b%5E2-4ac\" should not be a negative number.
\n" ); document.write( "
\n" ); document.write( " First, we need to compute the discriminant \"b%5E2-4ac\": \"b%5E2-4ac=%28-12%29%5E2-4%2A1%2A-28=256\".
\n" ); document.write( "
\n" ); document.write( " Discriminant d=256 is greater than zero. That means that there are two solutions: \"+x%5B12%5D+=+%28--12%2B-sqrt%28+256+%29%29%2F2%5Ca\".
\n" ); document.write( "
\n" ); document.write( " \"d%5B1%5D+=+%28-%28-12%29%2Bsqrt%28+256+%29%29%2F2%5C1+=+14\"
\n" ); document.write( " \"d%5B2%5D+=+%28-%28-12%29-sqrt%28+256+%29%29%2F2%5C1+=+-2\"
\n" ); document.write( "
\n" ); document.write( " Quadratic expression \"1d%5E2%2B-12d%2B-28\" can be factored:
\n" ); document.write( " \"1d%5E2%2B-12d%2B-28+=+1%28d-14%29%2A%28d--2%29\"
\n" ); document.write( " Again, the answer is: 14, -2.\n" ); document.write( "Here's your graph:
\n" ); document.write( "\"graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-12%2Ax%2B-28+%29\"

\n" ); document.write( "\n" ); document.write( "The answer must be positive so the answer is 14.
\n" ); document.write( "d=14
\n" ); document.write( "d-12=2
\n" ); document.write( "
\n" );